1.

A metal oxide has the formula M_(2)O_(3). It can be reduced by H_(2) to free metal and water. 0.1596g of M_(2)O_(3) required 6 mg of H_(2) for complete reduction. The atomic mass of the metal is:

Answer»

27.9
79.8
55.8
159.8

Solution :`M_(2)O_(3)+3H_(2)rarr2M+3H_(2)O`
`(2x+48)G 6G`
x=Atomic mass of METAL
`because` 0.006 g `H_(2)` reduces 0.1596g `M_(2)_(3)`.
`therefore 6g H_(2)` will reduce `(0.1596)/(0.006)xx6 g M_(2)O_(3)=159.6M_(2)O_(3)`
`2x+48=159.6`
`2x=11.6`
`x=55.8`.


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